Find the length of the minor axis of the ellipse 4x² + 9y² = 144 ?

CONCEPT:

The following are the properties of a horizontal ellipse \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) where 0 < b < a:

• Its centre is (0, 0)
• Its vertices are (- a, 0) and (a, 0)
• Its foci are (- ae, 0) and (ae, 0)
• Length of the major axis is 2a
• Length of the minor axis is 2b
• Equation of major axis is y = 0
• Equation of minor axis is x = 0
• Length of the latus rectum is given by \(\frac{2b^2}{a}\)
• Eccentricity is given by \(e = \frac{\sqrt{a^2-b^2}}{a}\)

CALCULATION:

Given: Equation of ellipse is 4x2 + 9y2 = 144

The given equation of ellipse can be re-written as \(\frac{{{x^2}}}{{{36}}} + \frac{{{y^2}}}{{{16}}} = 1\)

As we can see that, the given ellipse is a horizontal ellipse.

So, by comparing the given equation of an ellipse with \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) we get,

⇒ a2 = 36 and b2 = 16

⇒ b = 4

As we know, the length of the minor axis of an ellipse is given by 2b

So, the length of the minor axis is 2 ×  4 = 8 units

Hence, option B is the correct answer.